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a/ ĐKXĐ: ....
\(\Leftrightarrow x^2-8x+16+x+14-6\sqrt{x+5}=0\)
\(\Leftrightarrow\left(x-4\right)^2+\frac{\left(x+14\right)^2-36\left(x+5\right)}{x+14+6\sqrt{x+5}}=0\)
\(\Leftrightarrow\left(x-4\right)^2+\frac{x^2-8x+16}{x+14+6\sqrt{x+5}}=0\)
\(\Leftrightarrow\left(x-4\right)^2\left(1+\frac{1}{x+14+6\sqrt{x+5}}\right)=0\)
2/
\(A=\frac{5x}{2}+\frac{2}{5x}+\frac{7y}{2}+\frac{8}{7y}+\frac{1}{2}\left(x+y\right)\)
\(A\ge2\sqrt{\frac{10x}{10x}}+2\sqrt{\frac{56y}{14y}}+\frac{1}{2}.\frac{34}{35}=\frac{227}{35}\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x=\frac{2}{5}\\y=\frac{4}{7}\end{matrix}\right.\)
1.
\(PT\Leftrightarrow\left(x-4\right)^2+\left(\sqrt{x+5}-3\right)^2=0\left(x\ge-5\right)\)
\(\Leftrightarrow x-4=\sqrt{x+5}-3=0\Leftrightarrow x=4\).
\(P=3x+\dfrac{12}{x}+y+\dfrac{16}{y}+2\left(x+y\right)\ge2\sqrt{3x.\dfrac{12}{x}}+2\sqrt{y.\dfrac{16}{y}}+2.6=32\)
\(\Rightarrow P_{min}=32\) khi \(\left\{{}\begin{matrix}3x=\dfrac{12}{x}\\y=\dfrac{16}{y}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)
\(\sum\dfrac{x^4y}{x^2+1}=\sum\dfrac{x^3.\dfrac{1}{z}}{x^2+xyz}=\sum\dfrac{x^2}{z\left(x+yz\right)}=\sum\dfrac{x^2}{xz+1}\)
Áp dụng bất đẳng thức cauchy-schwarz:
\(Vt=\sum\dfrac{x^2}{xz+1}\ge\dfrac{\left(x+y+z\right)^2}{xy+yz+xz+3}\)
mà theo AM-GM: \(xy+yz+xz\ge3\sqrt[3]{x^2y^2z^2}=3\)
hay \(3\le xy+yz+xz\)
do đó \(VT\ge\dfrac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)}\ge\dfrac{3\left(xy+yz+zx\right)}{2\left(xy+yz+xz\right)}=\dfrac{3}{2}\)
Dấu = xảy ra khi x=y=z=1
P/s: Câu này khoai
Bài 1 :
Ta có : \(\dfrac{1}{3a^2+b^2}+\dfrac{2}{b^2+3ab}=\dfrac{1}{3a^2+b^2}+\dfrac{4}{2b^2+6ab}\)
Theo BĐT Cô - Si dưới dạng engel ta có :
\(\dfrac{1}{3a^2+b^2}+\dfrac{4}{2b^2+6ab}\ge\dfrac{\left(1+2\right)^2}{3a^2+6ab+3b^2}=\dfrac{9}{3\left(a+b\right)^2}=\dfrac{9}{3.1}=3\)
Dấu \("="\) xảy ra khi : \(a=b=\dfrac{1}{2}\)
Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel ta có :
\(S=\frac{1}{x}+\frac{1}{4y}+\frac{1}{16z}=\frac{1}{x}+\frac{\frac{1}{4}}{y}+\frac{\frac{1}{16}}{z}\ge\frac{\left(1+\frac{1}{2}+\frac{1}{4}\right)^2}{x+y+z}=\frac{\frac{49}{16}}{1}=\frac{49}{16}\)
Đẳng thức xảy ra <=> \(\hept{\begin{cases}x=\frac{16}{21}\\y=\frac{4}{21}\\z=\frac{1}{21}\end{cases}}\). Vậy GTNN của S = 49/16
1. Vì x, y, z > 0
\(xy+yz+zx\ge2xyz\)
\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge2\)
Suy ra:
\(\dfrac{1}{x}\ge1-\dfrac{1}{y}+1-\dfrac{1}{z}=\dfrac{y-1}{y}+\dfrac{z-1}{z}\ge2\sqrt{\dfrac{\left(y-1\right)\left(z-1\right)}{yz}}\). (1)
Tương tự \(\dfrac{1}{y}\ge2\sqrt{\dfrac{\left(z-1\right)\left(x-1\right)}{zx}}\) (2)
và \(\dfrac{1}{z}\ge2\sqrt{\dfrac{\left(x-1\right)\left(y-1\right)}{xy}}\) (3)
Nhân (1), (2), (3) với nhau theo vế ta được
\(\dfrac{1}{xyz}\ge\dfrac{8\left(x-1\right)\left(y-1\right)\left(z-1\right)}{xyz}\)
\(\Leftrightarrow\left(x-1\right)\left(y-1\right)\left(z-1\right)\le\dfrac{1}{8}\)
Đẳng thức xảy ra \(\Leftrightarrow x=y=z=\dfrac{3}{2}\)
Không mặn mà với số này cho lắm
\(A=\dfrac{5}{2}x+\dfrac{2}{5x}+\dfrac{7}{2}y+\dfrac{8}{7y}+\dfrac{1}{2}\left(x+y\right)\)
\(A\ge2\sqrt{\dfrac{5}{2}x.\dfrac{2}{5x}}+2\sqrt{\dfrac{7}{2}y.\dfrac{8}{7y}}+\dfrac{1}{2}.\dfrac{34}{35}\)
\(A\ge2+4+\dfrac{17}{35}=\dfrac{227}{35}\)
GTNN là \(\dfrac{227}{35}\Leftrightarrow\)\(\left\{{}\begin{matrix}x=\dfrac{2}{5}\\y=\dfrac{4}{7}\end{matrix}\right.\)